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Question: In ∆ABC, a = 5, b = 4, and cos (A – B) = \(\frac { 31 } { 32 }\), then side c is –...

In ∆ABC, a = 5, b = 4, and cos (A – B) = 3132\frac { 31 } { 32 }, then side c is –

A

6

B

7

C

9

D

None of these

Answer

6

Explanation

Solution

tan =1cos(AB)1+cos(AB)\sqrt { \frac { 1 - \cos ( \mathrm { A } - \mathrm { B } ) } { 1 + \cos ( \mathrm { A } - \mathrm { B } ) } }

= 1(3132)1+(3132)\sqrt { \frac { 1 - \left( \frac { 31 } { 32 } \right) } { 1 + \left( \frac { 31 } { 32 } \right) } }= 163\frac { 1 } { \sqrt { 63 } }

aba+b\frac { a - b } { a + b } cot = 163\frac { 1 } { \sqrt { 63 } }

19\frac { 1 } { 9 } cot = 163\frac { 1 } { \sqrt { 63 } }

⇒ tan = 73\frac { \sqrt { 7 } } { 3 }

Now cos C = 1tan2C/21+tan2C/2\frac { 1 - \tan ^ { 2 } \mathrm { C } / 2 } { 1 + \tan ^ { 2 } \mathrm { C } / 2 } ⇒ cos C = 17/91+7/9\frac { 1 - 7 / 9 } { 1 + 7 / 9 }= 18\frac { 1 } { 8 }

Q c2 = a2 + b2 – 2ab cos C

c2 = 25 + 16 – 40 × 18\frac { 1 } { 8 } = 36

⇒ c = 6.